Answer:
![EF=4\sqrt{3}](https://tex.z-dn.net/?f=EF%3D4%5Csqrt%7B3%7D)
Step-by-step explanation:
In rectangle ABCD, AB = 6, BC = 8, and DE = DF.
ΔDEF is one-fourth the area of rectangle ABCD.
We want to determine the length of EF.
First, we can find the area of the rectangle. Since the length AB and width BC measures 6 by 8, the area of the rectangle is:
![A_{\text{rect}}=8(6)=48\text{ cm}^2](https://tex.z-dn.net/?f=A_%7B%5Ctext%7Brect%7D%7D%3D8%286%29%3D48%5Ctext%7B%20cm%7D%5E2)
The area of the triangle is 1/4 of this. Therefore:
![\displaystyle A_{\text{tri}}=\frac{1}{4}(48)=12\text{ cm}^2](https://tex.z-dn.net/?f=%5Cdisplaystyle%20A_%7B%5Ctext%7Btri%7D%7D%3D%5Cfrac%7B1%7D%7B4%7D%2848%29%3D12%5Ctext%7B%20cm%7D%5E2)
The area of a triangle is half of its base times its height. The base and height of the triangle is DE and DF. Therefore:
![\displaystyle 12=\frac{1}{2}(DE)(DF)](https://tex.z-dn.net/?f=%5Cdisplaystyle%2012%3D%5Cfrac%7B1%7D%7B2%7D%28DE%29%28DF%29)
Since DE = DF:
![24=DF^2](https://tex.z-dn.net/?f=24%3DDF%5E2)
Thus:
![DF=\sqrt{24}=\sqrt{4\cdot 6}=2\sqrt{6}=DE](https://tex.z-dn.net/?f=DF%3D%5Csqrt%7B24%7D%3D%5Csqrt%7B4%5Ccdot%206%7D%3D2%5Csqrt%7B6%7D%3DDE)
Since ABCD is a rectangle, ∠D is a right angle. Then by the Pythagorean Theorem:
![(DE)^2+(DF)^2=(EF)^2](https://tex.z-dn.net/?f=%28DE%29%5E2%2B%28DF%29%5E2%3D%28EF%29%5E2)
Therefore:
![(2\sqrt6)^2+(2\sqrt6)^2=EF^2](https://tex.z-dn.net/?f=%282%5Csqrt6%29%5E2%2B%282%5Csqrt6%29%5E2%3DEF%5E2)
Square:
![24+24=EF^2](https://tex.z-dn.net/?f=24%2B24%3DEF%5E2)
Add:
![EF^2=48](https://tex.z-dn.net/?f=EF%5E2%3D48)
And finally, we can take the square root of both sides:
![EF=\sqrt{48}=\sqrt{16\cdot 3}=4\sqrt{3}](https://tex.z-dn.net/?f=EF%3D%5Csqrt%7B48%7D%3D%5Csqrt%7B16%5Ccdot%203%7D%3D4%5Csqrt%7B3%7D)
Answer:
a
Step-by-step explanation:
Answer:
x=85
Step-by-step explanation:
Hope this helps you :))
(Hope it doesn't get reported becuase it "doesn't have an explanation of how it's right") Have a blessed day ^-^
the answer for x is 76.4. the way to find it is by dividing 458.4 by 6 to find x s value.
Answer: There is no polygon with the sum of the measures of the interior angles of a polygon is 1920°.
Step-by-step explanation:
The sum of the measures of interior angles of a polygon with
sides is given by:-
![(n-2)\times180^{\circ}](https://tex.z-dn.net/?f=%28n-2%29%5Ctimes180%5E%7B%5Ccirc%7D)
Given: The sum of the measures of the interior angles of a polygon is 1920°.
i.e. ![(n-2)\times180^{\circ}=1920^{\circ}](https://tex.z-dn.net/?f=%28n-2%29%5Ctimes180%5E%7B%5Ccirc%7D%3D1920%5E%7B%5Ccirc%7D)
![n-2=\dfrac{1920}{180}\\\\ n= \dfrac{1920}{180}+2\\\\ n=\dfrac{2280}{180}=12.67](https://tex.z-dn.net/?f=n-2%3D%5Cdfrac%7B1920%7D%7B180%7D%5C%5C%5C%5C%20n%3D%20%5Cdfrac%7B1920%7D%7B180%7D%2B2%5C%5C%5C%5C%20n%3D%5Cdfrac%7B2280%7D%7B180%7D%3D12.67)
But number of sides cannot be in decimal.
Hence, there is no polygon with the sum of the measures of the interior angles of a polygon is 1920°.